Y RAJYALAKSHMI11 Years agocos2x sin4x/[cos4x(1+cos2 2x)] can be written as tanx – 16cos3x sin3x / [cos4x(3+cos4x)] ∫tanx dx = sec2xput cos4x/(3+cos4x) = tthen – 16cos3xsin3x/(3+cos4x)2 dx = dt∫– 16cos3x sin3x / [cos4x(3+cos4x)]dx = ∫1/t dt = log t = log [cos4x/(3+cos4x)] = 4log(cosx) – log(3 + cos4x) So, the integral of given function is sec2x + 4log(cosx) – log(3 + cos4x) + C